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Calculus problem

\[\large \dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2+t^2}} \, dt = \, ? \]

Note by Starwar Clone
3 months ago

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@Pi Han Goh @Guillermo Templado help will be welcomed . Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone Tomorrow, I rewiewed this, and if it's not solved. I'll try to solve it, I think I'm not going to have a lot problemms.. we'll see... Guillermo Templado · 2 months, 2 weeks ago

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@Starwar Clone I agree with Pi Hang Goh, you can also make the change \(t = x\Sh(u)\) solve , the integral, and later derivate Guillermo Templado · 2 months, 2 weeks ago

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@Guillermo Templado I mean \( t = x \cdot \text{Sh u}\). I have some problems with my comments.. Guillermo Templado · 2 months, 2 weeks ago

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Hint: Fundamental Theorem of Calculus. Pi Han Goh · 3 months ago

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@Pi Han Goh yeah but i am still not able to get the required answer will you please post the solution to this question... Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on. Pi Han Goh · 2 months, 2 weeks ago

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@Pi Han Goh i am getting the answer \[ \dfrac{3x^2}{\sqrt{x^2 + x^12 }} \] it is x^12 .. in the denominator Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone Split the integral into \( \int_0^{x^3} - \int_0^{t^2} \). For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you? Pi Han Goh · 2 months, 2 weeks ago

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@Pi Han Goh Yeah that is enough Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone Is the answer correct? Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone Make the change I have said to you... It must solve it quickly.. or use Fundamental theorem of calculus Guillermo Templado · 2 months, 2 weeks ago

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@Guillermo Templado can you provide the final answer , i am really stuck on this ? Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone you have to develop the steps... Guillermo Templado · 2 months, 2 weeks ago

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@Guillermo Templado do you mean that i really need to integrate the above expression first? Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone It's not necessary, but if you make the change \(t = x \cdot \text{ Sh u} \) with \(x\) constant, and use \(1 + Sh^2(u) = Ch^2(u)\) and \(dt = x \cdot \text{Ch u} \space du\), the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise.... Guillermo Templado · 2 months, 2 weeks ago

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@Guillermo Templado thanks man , appreciate the help Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone Ok, come on, please tell me if I'm wrong, I'm a person and due to this I make a lot of(too many) mistakes...

Proof 1 (attempt 1).-

\[\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} \left( \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt - \int_{0}^{t^2} \dfrac 1{\sqrt{x^2 + t^2}} \, dt \right) =\] Let's call \(g(x) = x^3\) and \(f(x) = \int_{0}^{x} \dfrac 1{\sqrt{x^2 + t^2}} \, dt\) Then, \[\dfrac d{dx} \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} f(g(x)) = \dfrac d{dx} f(x^3) = 3x^2 \cdot ...\] Sorry, I'm getting a knot for myself...

Proof 2.-

\[\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \space (*)\] Let's make the change \( t = x \cdot \text{ Sh u}\) , \(x\) constant, \(dt = x \cdot \text{Ch u} \space du\) \[ (*) = \dfrac d{dx} \int_{Sh^{-1}(x \cdot \text{ Sh}^{2}u)}^{Sh^{-1} x^2} \, \pm \space du =...\] To be continued Guillermo Templado · 2 months, 2 weeks ago

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@Guillermo Templado Hmm it's getting better Starwar Clone · 2 months, 2 weeks ago

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@Starwar Clone I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also if \(x >0\), \(x = 0\) or \(x < 0\) to concrete the ending. This question is giving me questions and suggestions... For example, does \(\int_0^{t} \space dt\) make some sense?If the answer is yes, Is it \(t\)?why?... I have another proof making the change \(u = \sqrt{x^2 + t^2} + t\)... Even we can think first this question without the derivative \(\frac{d}{dx}\), I mean given \(x \) possible concrete constants, \(x =1\), \(x =2\), \(x = 100\)... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare... Guillermo Templado · 2 months, 2 weeks ago

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